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2. Criteria For The Load Flow
Study
1. Collection of all data about the Nablus network
including the one-line diagram, information about the
power stations, transformers, transmission lines and
loads.
2. Investigating the problems from which the Nablus
network suffers in max., min load & fault condition
3. Applying method of reactive power compensation to
improve the operation (using tap changer transformer,
capacitor bank)
4. Performing the economical analysis of the saving
achieved by the implementation of reactive power
compensation

3. CH:2 Existing System
The present load in the West Bank is supplied from
several points within the IEC network. There are supply
points at
22 KV supplying Qalailya and Tulkarm
33 KV feeders from Beisan (in
Israel)supplying Jenin ,Tubas and
Nablus as well as feeders from M.
Afraym and feeders from portable
substations.

4. Load Forcast

5. 3.1: Element Of The Network
A. Sources
. Generators are one of the essential components of the power
systems. Synchronous generators are widely used in power
systems.
. Nablus are fed from 3 connection point by Israel Electrical
Company (IEC), at 33KV.
1.Asker (odeleh & Almeslekh) →→→30MVA namely
2.Quseen →→→18MVA namely
3.Innab →→→5MVA namely

6. B. Transformers & load
• There are two type of Transformer (power &
distribution Transformer ) which is ∆-y connected
• So, the source always balanced without looking to the
load since the P.T is ∆-y ground connected
• The distribution Transformer capacity & its voltage
are summarized in table (1)

7. • There are also 3 power Transformer with (10MVA
capacity,33/6.6KV& %Tap=±12 ;for 17 taps with1.5% for each step
changer)
• Note the impedance of each transformer are determined by the
typical value at ETAP program
• The loads will be seen later in the table of L.F of transformer

8. C.Transmission Line
• There are two type of conductor
1.O.H lines →→→ACSR
2.cables →→→ cu XLPE
which have these rating as in table (2)
. In Quseen they use double T.L (3*95mm2ACSR) for the
transmission of power since the current is above 400A in max
load

9. 3.2 : Electrical Problem In The Network
1. High drop voltage
2. Low power factor
3. Low load factor at most of distribution transformer which
reduce the efficiency of transformer & so increase the losses
(the electrical losses in distribution network must not exceed
8-10 % from total active power)
4. Exceeding the permissible capacity of each connection point
MVA

10. Remedy
1. High drop voltage & low power factor problem
• It is important to keep the power factor above 0.92 on the
distribution transformer so as to minimize the electrical
losses in the network & do not paying penalties
• For max load the voltage of buses must rise to
(1.05Vnom<= Vbus <=1.1Vnom) as we can to decrease the
current & so to decrease the losses
• The first step for this improvement is done by using the taps ,
if not enough it can solve by adding capacitor banks

11. 2. Low load factor problem
• Re-arrange the distribution transformer (if it can be) to
increase the load factor
• LF=[.65-.75] give the max. efficiency distribution transformer
• The engineers choose the transformer in distribution network
with load factor [.45-.55] expressed to the growth of the load
by years
3. Capacity problem
• For the fourth problem there is a study to get another
connection point or to move the connection to another point
as in Qussen

12. Taps
• Almost all transformers provide taps on windings to adjust
the ratio of transformation by changing taps down when we
need to raise the voltages up and vice versa
• There are two types of transformer taps:
1. Tap changing without loads (fixed tap) changer on either side
or both sides of transformers
2. Tap changing under load (LTC)

13. capacitor banks
• Shunt capacitor banks is very important method of controlling
voltage at the buses at both transmission and distribution
levels along lines or at substation and load .
• Essentially capacitor is a means of supplying mega-vars
(MVAR) at the point of installation.
• Capacitor banks may be permanently connected, or regulators
• Switching may be manually or automatically controlled either
by time clock or in response to voltage or reactive - power
requirement
• capacitor reduces the line current necessary to supply the load
and reduce the voltage drop in the line as the power factor is
improved

14. 4.1: Etap Power Station Program
• It is a load flow program which can simulink the
power system receiving the input data (source
,transformer ,T.L & loads) as One Line Diagram
schematic And results output report that includes
bus voltage , branch losses , load factors power
factors …etc.
• It is also able to do the Fault analysis .. Harmonic
analysis .. Transient stability analysis.

15. 4.2: Simulation For Max. Load Case
This step done by the following criteria
1. drawing the one line diagram (source ,transformer
T.L, buses & loads)
2. entering R&X in Ω or (Ω /any unit of length) & its
length. note (Y) value is not important since the T.L
is short (L<80Km)
3. entering the typical value (X/R & %Z) for each
transformer
4. entering the rated voltage for each bus
5. entering the actual MVA & P.F for each load
6. entering the source as a swing bus, for load flow
studies a swing power grid will take up the slack of
the power flows in the system, i.e., the voltage
magnitude and angle of the power grid terminals
will remain at the specified operating values ( V & δ
are given ,P & Q are unknown)
7. run the load flow analysis to get the output result

16. A: Max. load case results without
improvement
• The total demand for Qussen
Swing bus P= 20.27MW Q=14.846MVAr S=24.929MVA pf= 80.33 lagging
∆P=1.06 ∆Q=2.555 I=436A
∆P%=1.06/20.27=5.456%
• The total demand for Innab
Swing bus P= 5.702MW Q=3.741MVAr S=6.82MVA pf= 83.61 lagging
∆P=.075 ∆Q=.307 I=119A
∆P%=.075/5.695=1.31%
B: Qussen-with tap changer improvement
Swing bus P= 19.841MW Q=14.318MVAr S=24.467MVA pf= 81.09 lagging
∆P=.920 ∆Q=2.026 I=428A
∆P%=.92/19.841=4.63%
Method of iteration: Newton Raphson method Number of Iterations: 3

17. L.F for Quseen

18. L.F for Innab

19. P.F for Quseen

20. P.F for Quseen

21. P.F for Innab

22. V% for Quseen

23. V% for Quseen

24. V% for Innab

25. Problems In The Network
• we notice # of problems:
1. low load factor (L.F<.45) for the most of transformer
2. high load factor (L.F>1) for some transformer
{T82,T103 in Quseen}
3. The P.F for all buses are low (P.F<.92) except
{bus.33,34 in Quseen & bus 25,27,29,54 in Innab}
4. %V does not lies between(1.05Vnom-1.1Vnom) for any
bus
5. considerable losses in Quseen (∆P%=5.456)

26. V% for Quseen with Taps

27. V% for Quseen with Taps

28. V% for Quseen with Taps

29. For this case we notice the following result
1. there is a small increase in the P.F
2. 36% of buses lies between(1.05Vnom-
1.1Vnom) & the other is >95% Vnom
3. P.F of the swing bus increase from 80.33 to
81.09
4. The current decrease from 436A to 428 A
5. the losses decrease in Quseen .826% from the
original case
6. There is a saving in the capacity of .5MVA

30. C:Max.load with capacitor improvement
* Capacitor bank are used to solve P.F problem &
its penalties, we put these capacitor bank at the
load side (0.4Kv side)
* Qc=pold(tancos-1p.fold-tancos-1p.fnew)
Where standard capacitor are:-
0.4Kv→→→25,40,60,100KVAr
6.6or11Kv→3,6MVAr

31. 1.Quseen
Qc=19.841*(tancos-1.8109-tancos-1.92)
=5.866MVAr
Qcact=5.476MVAr for P.F=.9202
2.Innab
Qc=5.702(tancos-1.8361-tancos-1.92)
=1.312MVAr
Qcact=1.3MVAr for P.F=.9221
for each load to
we use suitable rated capacitor bank
*
rise its p.f above .92,so to increase the overall p.f of
swing bus

32. PF for Quseen with cap

33. PF for Quseen with cap

34. PF for Inab
with cap

35. 4.2: Simulation For Min. Load Case
A : Min. load case results without improvement
The total demand for Qussen
Swing bus P= 7.721MW Q=5.21MVAr
S=9.348MVA pf= 82.59 lagging
∆P=.152 ∆Q=.354 I=164A
∆P%=1.96%
The total demand for Innab
Swing bus P= 2.262MW Q=1.421MVAr
S=2.672MVA pf= 84.69 lagging
∆P=.012 ∆Q=.047 I=47A
∆P%=.53%

36. V% for Quseen without tap

37. V% for Quseen without tap

38. B: Qussen-with tap changer improvement
Swing bus P= 7.709MW Q=5.2MVAr
S=9.32MVA pf= 82.72 lagging
∆P=.141 ∆Q=.32 I=163A
∆P%=1.82%
At this case half turn of tap changer are used
*
to increase the voltage of the bus (vbus>=vnom)
[only Quseen region have under this value
Vbus=.95-.98.5Vnom] .tap changer have affect
to increase the voltage but less affect on p.f.

39. V% for Quseen with tap

40. V% for Quseen with tap

41. C: Min load using capacitor
1.Quseen
Qc=7.709*(tancos-1.8272-tanco-1.92)
=1.952MVAr
Qcact=1.995 MVAr for P.F=.9224
2.Innab
Qc=2.262(tancos-1.8469-tancos-1.92)
=.456MVAr
Qcact=.547 MVAr for P.F=.9209

42. PF for Quseen without cap

43. PF for Quseen without cap

44. PF for Inab
without cap

45. PF for Inab
with cap

46. PF for Quseen with cap

47. PF for Quseen with cap

48. At min load less capacitor bank are used to
*
rise the p.f at the load & so the overall p.f. the
losses in the network are become very low
since the currents is reduced .
Some of these capacitor bank are used at
*
max &min which is called fixed capacitor bank.
And other capacitor which only used at max or
at min are called regulated one. regulated
capacitor bank are more expensive than fixed
since it need to controller for use.

49. Changing of the switch gear & connection point
simultaneously
 change the switch gear from 33/6.6KV to 33/11KV except Jumblat
region
• Jumblat region will kept as it is to exploits the distribution
transformer which have two level voltage at primary side (11,
6.6KV/.4KV) at Jumblat with transformer have only 6.6/.4KV side at
other places ( East & West Mojeer aldeenregion )
• This operation will save the price of a new transformer with 11/.4KV
 The change of position of the connection point is from Qussen to
sarrah with 3Km double T.L

50.  The change is starting from replacing the distribution
transformer of 6.6/.4KV in East & West Mojeer aldeen
region to 11,6.6/.4KV(from Jumblat & East part region
from Nablus)
 This step also taken some case of the L.F distribution
rearrangement which are sumerize at the next table

51. from
New capacity
L.F
capacity
T#
Jumblat
630
58.6
630
62
Jumblat
400
29.5
630
63
Jumblat
630
69.1
630
64
Jumblat
630
91.6
400
68
East part
250
36.8
250
72
East part
160
47.3
160
74
East part
250
59.4
250
75
Jumblat
400
38.2
400
76
Jumblat
400
29.2
400
77
Jumblat
84
42.8
84
84
Jumblat
630
26.5
630
90
Jumblat
400
44.9
400
97
Jumblat
630
40.7
630
99
Jumblat
630
40.7
630
100
Jumblat
630
52.2
630
101
Jumblat
400
105.7
300
103
Jumblat
400
91.2
400
104
East part
1000
1000
108

52.  The transformer at Jumblat region which exchange are
T25,T26,T27,T28,T30,T38,T42,T45,T46,T47,T49,T50,T51
&T61
630KVA→→ T25,T26,T30,T42,T45&T50
400KVA→→ T27,T28,T39,T46,T47,T49, T51&T61
 the transformer of Jumblat are back as its default
capacity except
1)T39(400KVA) of L.F=18.3 with T103(300KVA)
2)exchange T47(400KVA) of L.F=80.4% with T25(630KVA)
of L.F=7.7%
The result of this case with taps:-
Swing bus P= 19.647MW Q=14.172MVAr S=24.225MVA
pf= 81.1 lagging
∆P=0.727 ∆Q=0.881 I=424A
∆P%=0.727/19.647=3.7%

53. From ETAP result we notice that:-
1. the current in The main T.L at Quseen is decreased in
widely range from 400 to 35A approximately due to change of
the position of connection point to Sarah.
at Sarah the power distribute directly for more branches. The
decrease in current will decrease the losses in The main T.L
(at the old case)
2. the change from 6.6 to 11KV also decrease the current in
the branches & so this mean decreasing in the losses
3. the voltage is slowly decreased at Quseen busses region
since the supply is exchange
4. the voltage is slowly increased at the region which the
switch gear is change

54. Quseen,old
case

58.  The first economical study is using capacitor bank &
this study followed by this criteria:-
∆∆P=∆Pbefore,cap - ∆Pafter,cap
∆∆P:saving in real power losses
∆Pbefore,cap : real power losses before adding capacitor
∆Pafter,cap : real power losses after adding capacitor
Z∆p=∆∆p*T*140
Z∆p : annual saving in real power cost
T=8760(0.124+0.0001tmax)^2
T≈3500hour
140:cost per MWh($/MWh)
Kc=C*Qc
Kc:cost of capacitor
C:cost of capacitor per KVAr($/KVAr)
Qc: capacitor KVAr

59. knowing that the cost of the capacitor are:-
Fixed Cap=5$/Kvar
Regulated Cap=22$/Kvar
The type of capacitors used are regulated only since the loads are vary every
time , increase by years & may be decrease under min. so the P.F become
leading & this will damage the transformer . also the P.F correction range is
considerable to use regulated capacitor.
Zc=0.22*Kc
Zc:annual capacitor running cost
.22: maintenance & life time of capacitor (depreciation factor)
∆Z=Z∆p-Zc
∆Z: annual saving
Saving of penalties=1% from the total bill for every 1% p.f <92%
∆Zt=∆Z+ Saving of penalties
∆Zt: total annual saving
S.P.B.P=investment(capacitors initial cost)/ total annual saving
S.P.B.P < 2year →→→project is visible
S.P.B.P > 2year →→→project is not visible

60. 1)Quseen
∆∆P=∆Pbefore,cap-∆Pafter,cap
∆∆P=0.92-0.702=0.218
Z∆p=∆∆p*T*140
Z∆p=0.218*3500*140=106820
Kc=C*Qc
Kc=5.476*10^3*22=120472
Zc=0.22*Kc=26503.84
∆Z=Z∆p-Zc=106820-26503.84
=80316
Total Bill=19.841*10^3*3500=69443500kwh
Saving of penalties=1%*69443500 (92-81.09)*140*10^-3=1060680
∆Zt=80316.16+1060680=1140996(This is the annual saving)

61. 2)Innab
∆∆P=∆Pbefore-∆Pafter
∆∆P=0.075-0.061=0.014
Z∆p=∆∆p*T*140
Z∆p=0.014*3500*140=6860
Kc=C*Qc
Kc=1.3*10^3*22=28600
Zc=0.22*Kc=6292
∆Z=Z∆p-Zc=6860-6292
=568
Total Bill=5.702*10^3*3500=19957000
Saving of penalties=1%*19957000 (92-83.61)*140*10^-3
= 234415
∆Zt=568+234415=234983 (This is the annual saving)

62. S.P.B.P=total investment/total annual saving
=(120472+28600) / (1140996+234983)
=.1083 year
=1.296 month
 The second economical study is changing of switch gear & the
connection point.this step will make saving in power without paying
money
Annual saving=saving in power*3500h*140$/MW
=Pold-Pnew *3500h*140
=(19.841-19.647) *3500*140
=95060$/year